GEEZ that list is huge! And I can hardly imagine how long it took to put it all together! More great work guys. Give yourselves a pat on the back (or a cookie, if you prefer ^^) (In fact... *gives cookies to X-act and peterko*)

Apparently, all IVs have an equal chance of being selected when chosen at random, so if you, for some reason, want to have an HP IV of 24 for whatever reason, then just replace all mentions of 31 HP with 24 HP on the single "flawless" HP section.

In other words (and correct me if I'm wrong here), it doesn't matter how high you want the IV to be, if you're going for a fixed number and the parents have that same fixed number, these are the probabilities.

That'd be passing 4 IVs down, so no. The baby's random IVs might potentially match, giving it the appearance that 4 were down, so it's possible to get the IVs you're looking for, but not in the way that you described.

So would it still be good to put both good IV parents then even if you only pass down three? And is there a bigger chance the offspring will have better IVs if both parents have good IVs? Like 2 Bagons have 23 and 25 attack. Is there a chance the offspring can have a better attack?

I call it "waiting for a random high (max) IV", it is quite worth doing it with HP as
- HP has the highest probability to be chosen at random, the lowest probability to be passed through breeding
- defense has a lower probability to be chosen at random than HP but a higher probability to be passed through breeding
- attack/special attack/special defense/speed -> they will be passed very often so don´t try to wait for a random high on these

it should be clear what you should aim for when breeding for a particular IV combination, just take a look at the numbers

so ideally, after 158 eggs you have your max HP/at/speed pkmn, but what if you want both defs to be 20+ or 25+? you would try to get both parents with 31hp/at/speed (another 62 eggs) and wait for random defs to be high (now you get 31 hp/at/speed every 39 eggs), but they are passed really often so it would take too much time...

so why not aim for 31at/sD/speed? the easiest combination to obtain -> in 84 eggs (4+17+9+54), to have both parents with 31at/sD/speed in just 113 eggs (compared to 220 in the first example), and now you get a 31at/sD/speed baby in every !16! eggs with a good chance to get a high random HP...which way would you go?

It was first realized when BobDoily presented some data that showed that out of the many Pokemon he bred, HP and Defense were being passed down less often than the other stats. I'm not sure what happened after that, but I think loadingNOW took a look at the ROM and then found the code responsible for this effect.

I have a max Atk/Spd Charmander and I'm breeding it with my 30 HP / 30 Atk / 31 Spd Ditto in hopes of passing the 30 HP and 31 Spd. Would I be better of breeding that Atk/Spd Char with an Adamant Ditto that has either Atk or Spd and another max stat (sadly I don't have a max Atk/Spd Ditto) then continue with this one and just hope I get lucky?

Of course, if I get a very good Defensive Chari, I suppose I could shoot for a lower HP.

I've always wondered how people breed these monster Pokes I'm seeing everywhere in the threads.

For example, say you have a parent with 31/x/31/x/x/x and you are breeding with a parent with no max IVs. It is still possible to pull out a third (or more) max IV stat as well as passing down the HP and Def, albeit somewhat rarer than the other parent having a max IV to contribute

Well, I'm trying to breed a Smeargle with 6 31's (yes, I've seen the probabilities, but I'd like to try anyway xD) and I've been trying to work out the best way to do so. I've now got several Smeargle's with 2 perfect IV's in Attack, Sp. Attack, Sp. Defense or Speed, and one with 3 perfect IV's in HP, Sp Attack and Speed.

Would it be best to first make one with 3 flawless in Attack/Sp. Attack/Sp. Defense/Speed and then try to make another triple flawless and breed those to get a quad flawless in "Attack/Sp. Attack/Sp. Defense/Speed" Which I can use to "wait for a max IV"
or to use the HP/Sp Attack/Speed one, with one with flawless Sp. Def and Attack to get quad with easily passable flawless immidiately (replacing any parent with 3 easily passable flawless if I get one.)

So in short:

31/x/x/31/x/31 + x/31/x/x/31/x = Better parent

or

First get parents with 3 easily passable IV's, then breed those to get quad flawless.

Also, maybe it'd be nice to mention the applet in the guide itself, as I only saw it when I read the posts.

The applet is mentioned in the breeding guide; not in Part III though.

As for your question:

The best way would be to have two parents whose IVs, when interlaced, become a set of six perfect IVs. So if you have an x/31/x/x/x/31, this is best bred with 31/x/31/31/31/x. If you have 31/x/x/31/31/x, this is best bred with x/31/31/x/x/31. So first it's better to actually breed for two such parents, and then breed two such parents together.

Say you end up with 31/x/x/31/31/x and x/31/31/x/x/31. Breeding these, you should eventually get a Pokemon with four 31 IVs. Switch this Pokemon with the appropriate parent and breed those... until you end up with a perfect 31/31/31/31/31/31 Pokemon.

Hmm, now, I'm breeding for Ponytas, and I'm aiming for 31/31/20+/x/20+/31. The parents are 31/31/25/x/x/31 and x/31/x/x/x/31. I already know the chances of getting another 31/31/x/x/x/31 are 1:62, but how do I calculate the chances to get at least 20+ at both defenses?

i know this is a pretty old topic, but i have a question about the actual calculation of some of these numbers. in particular this one:

31/31/x/x/x/31 + x/31/x/x/x/31 --> 31/31/x/x/x/31.

i seem to get 1/120 whenever i try and calculate it manually (rather than the stated 1/62). i must be doing something wrong, because 1/62 is on the chart here, and also on the applet.

ignoring random 31's for the time being.....

the way i see it is that the hp must be passed in the first turn, with a 1/12 chance. the second IV to be passed is chosen from 10 possibilities (now that HP is not selected), and must be the atk or spd of either parent, so 4/10. the third IV to be passed is chosen from 8 possibilities (now that HP and Def are not selected), and it must be whichever of atk/spd was not chosen in the 2nd step, so this probability is 2/8. thus the probability of thw three stats being passed is

1/12 x 4/10 x 2/8 = 1/120.

i know there are also random chances (eg two 31's are passed, and another 31 appears by chance) but surely these are small enough not to bring it back up to the full 1/62?????

i'd appreciate some clarification from the experts, along with (possibly?) an explanation/derivation.

i know this is a pretty old topic, but i have a question about the actual calculation of some of these numbers. in particular this one:

31/31/x/x/x/31 + x/31/x/x/x/31 --> 31/31/x/x/x/31.

i seem to get 1/120 whenever i try and calculate it manually (rather than the stated 1/62). i must be doing something wrong, because 1/62 is on the chart here, and also on the applet.

ignoring random 31's for the time being.....

the way i see it is that the hp must be passed in the first turn, with a 1/12 chance. the second IV to be passed is chosen from 10 possibilities (now that HP is not selected), and must be the atk or spd of either parent, so 4/10. the third IV to be passed is chosen from 8 possibilities (now that HP and Def are not selected), and it must be whichever of atk/spd was not chosen in the 2nd step, so this probability is 2/8. thus the probability of thw three stats being passed is

1/12 x 4/10 x 2/8 = 1/120.

i know there are also random chances (eg two 31's are passed, and another 31 appears by chance) but surely these are small enough not to bring it back up to the full 1/62?????

i'd appreciate some clarification from the experts, along with (possibly?) an explanation/derivation.